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Why don’t airports use jetways as an airlock to avoid having to repressurize the plane for every flight?
Seems like it could save money perhaps. Has this been considered in the past?

How would they load & unload freight & luggage?

Probably the fastest 4 answers I have seen...  This one was low hanging fruit, thanks for the easy pitch!

Can you imagine all the wailing babies in the vicinity of the airlock?

@MichaelHall Gotta let people get their rep up (:

Could you edit this question to explain how you think that might save money?

I believe there are quite a few misconceptions here:

When an aircraft is "pressurised", it means that at higher altitudes, the pressure inside the aircraft is higher than the pressure outside. At lower altitudes, the pressure is exactly the same inside and out.
Originally, aircraft weren't pressurised, and at higher altitudes the low pressure is a problem for the people on board (lack of oxygen, mostly), and for some hardware (due to the pressure differential).
So the pressure is artificially increased inside the aircraft beyond a given altitude (or more precisely, beyond a given outside pressure): when the aircraft takes off, the pressure goes down as the plane reaches higher altitude, until that pressure is reached, and from then on, the pressure outside continues to decrease while the pressure inside is maintained at that safe level.
So there's no reason to pressurise an aircraft on the ground. It has the same pressure inside as outside.
Even if it were useful, when on the ground, there are plenty of openings other than just the passenger door(s). Other doors are opened for catering and technical staff, cargo doors are open, and so on. So it would be impossible to maintain a different pressure in the aircraft and outside without completely breaking down all operations.
Also, having an "air lock" would require doors on each side. One door would be opened, people would enter the air lock, the door would be closed, the pressure would be changed, the other door would be opened, people would exit the airlock. Then the door would be closed again, pressure brought back to the other level, and the first door would be opened again before more people would be able to enter the lock. This would significantly hamper the flow of people embarking or disembarking.
Additionally, people are already annoyed enough by the changes in pressure on board planes which take quite a few minutes. Doing that in a few seconds to ensure relatively quick movement would just be horrible.
Finally, consider that many aircraft do not use jetways but stairs on the tarmac.

Great comment. Especially the point near the end. You could only change the pressure altitude in the airlock about 500 feet per minute (1000 ft/min at best). Even if you had an airlock big enough to accommodate the entire plane load of passengers, that would add about 15 minutes per plane load in DFW. Or, you would have to pressurize (depressurize in the case of DFW) the entire building.

Another point is that pressurization in flight is a continuous process, exchanging fresh outside air for cabin air that's full of exhaled CO2.  So you'd have to run the plane's pressurization system (or the airport's) on the ground as well, costing even more money.

Pressurizing an entire building isn't impossible, @DeanF. (I know, you didn't say that.) Look at all the stadiums that have air supported roofs*. There are loads of revolving doors to keep the pressure inside, and the ear pop while entering is very annoying, but they're used on a regular basis. *The "Hoosier Dome" (later renamed "RCA Dome") in Indianapolis, and the "Humphrey Dome" in Minneapolis are two that come immediately to mind.

@FreeMan - I never said pressurizing the entire building was impossible. I essentially said it was impractical. Remember, regardless of whether you pressurize the air around the passengers at the jetway or at the front door of the building, it would be a drastic pressure change depending on where the building was. DFW is roughly 500 feet Pressure Altitude compared to 8000 feet cabin pressure altitude. The pressure difference between the inside and outside of the domes you mentioned was minimal.

Apart from the practical implications, aircraft pressurise their cabins with the bleed air coming from the jet engines. This air is compressed by the engines and used for the combustions process. So there are no seperate air compressors. If you want to have the aircraft compressed at the ground, you will need to either keep the engines running or add equipment to the aircraft. The first thing is not beneficial and the latter will increase the fuel consumption of the aircraft.
Have a look at the following picture:

From the picture it can be seen that the pressurization level of the cabin equals the pressure level at 7000ft. The means that if you keep this pressure then on the ground there is an underpressure in the aircraft, aircraft doors and seals are not designed with this in mind. The aircraft door for example is designed such that it can't be opened mid flight since the pressurisation of the cabin pushes the door outwards in the seals. This is the reason why aircraft doors and emergency hatches first move inwards before opening to swining outside of the aircraft.
To summarize, aircraft pressurization is there for the comfort of the passengers and is therefore kept at a comfortable level 7000ft. If the aircraft flies lower then, the inside pressure equals the outside pressure. And seconly, aircraft are not designed for the cabin to be of a lower pressure than the outside of the aircraft.

Comfort and brain function/health.

The only reason planes are pressurized is to allow the passengers the ability to breathe at flight altitude as if they were on the ground. If the plane is still or already on the ground, there would be no purpose in pressurizing it.
Most metal bodied aircraft pressurize their cabins to about 8000 feet cabin pressure altitude. If the outside air is at a pressure altitude below 8000 feet, the engines are just supplying air conditioning and circulation to the cabin, not necessarily added pressure.
The average human can breathe perfectly well below 15000 feet. Some may feel mild discomfort above 10000 feet. The aircraft engines breathe air and eat fuel best at altitudes of two to three times that. The airlines save money by flying at altitudes where they get the best fuel efficiency. Unfortunately, our bodies are not efficient like engines at that altitude. We would pass out or pass away. The average human has between 6 and 60 seconds of useful consciousness at flight altitudes.
The way to save money is to reduce the self-loading-cargo’s reliance on oxygen. Putting all the passengers asleep while wearing positive pressure oxygen masks would eliminate the need to pressurize the entire cabin .

But that many oxygen masks, which would need to be inspected & maintained for the life of the aircraft, would probably be more expensive than building a pressurized hull.

Sooooooooooo, you’re saying just put them to sleep with no mask. OK.

By the way, jk.

Below around 8000' MSL aircraft pressurization is roughly equal to the outside air pressure.  Since most airports are below this elevation, there is no pressure differential once the airplane is on the ground, therefore no need or even possibility for an airlock.
It takes very little bleed air to pressurize when airborne so there isn't a significant cost in doing so.  Furthermore, jetways aren't sealed so the additional cost in equipment and procedures to implement this plan even if it were possible or desirable would far outweigh any savings.

Your first comment about the differential pressure being roughly equal below 8000 is inaccurate. I think you may have some misconceptions about how these systems work. These systems typically start pressurizing before or just after takeoff. The pressure at, say 5000 MSL after taking off from sea level, might be around 500 ft cabin altitude.

You are correct, I don't fly pressurized passenger planes.  I'm ex Navy.  The EA6B was ambient to 8k.  And the Caravan is of course unpressurized.

Ok, yeah I can imagine that military pressurization systems might be substantially different from pressurized passenger aircraft.

Thanks for pointing out that my answer may not be as universally correct as I hoped it would be.  I may edit later to incorporate your comments, but I think the gist of it is sufficient to discredit the idea of a need for airlocks.  Although maybe I wont bother now, because there are better answers posted.

No problem. Your overall point is definitely valid.

Besides the fact that this is unlikely to be kind on passengers to overpressurize, the passenger doors and the jetways aren't the only portal into the pressurized space.  The luggage going in the belly and the servicing carts coming in the rear would also have to be accommodated.  Trying to load and unload cargo through an airlock would slow things down a lot.
It's possible that such an effort would extend the service life of airframes, but I don't see how it would save any money.  New pressurization equipment is needed while on the ground, portals would need extra material to interface with airlocks (increasing weight), and you'd need more time on the ground to move passengers and cargo.  Together, those seem to all be extra costs, not money savers.

Pressurising the cabin of even the largest airliner really doesn't take much energy at all, not least because it is achieved via the ventilation system, so even if all the other answers were irrelevant, it still wouldn't be cost-effective.


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Can you identify these antenna/items on the belly of a Cessna 172?
Not sure what all these antenna are for. I have my checkride soon and want to make sure I can name everything that might be pointed out. Thank you.

You're wandering around a hangar. Why not find the aircraft's owner and ask them?

Why are you asking internet strangers about an airplane in which you are going to take a checkride? Ask your flight instructor. That is why you are paying them. If it is not the aircraft in which you are taking your checkride, the DPE is not going to ask you about it. And, a DPE is not going to ask you about items under the cowling on a PPL checkride (per your other question). If you are going for a checkride above PPL, you should not have to ask these questions. You should be able to do the research on your own (POH). If you don’t know your plane, you are not ready for a checkride

@Dean F. I have never seen antenna locations in a POH myself.  Wouldn't he need to look in a service manual?

@RyanMortensen - A service manual is a good call. The owner, CFI or A&P of that aircraft are an even better resources. Things might have been moved since 1977. Honestly, if he is going for a PPL or under checkride, the service manual may be beyond the scope of knowledge the DPE would expect. Since checking the antennae (visually and for looseness) is probably on the OPs preflight checklist, the CFI should have gone over which antennae those are in the first or second flight.

@DeanF. FYI. I did do my own research. I asked my CFI and A&P, as well as looked in the POH, but didn’t get a definitive answer, so I thought it would be helpful to ask internet strangers in case they have a wider scope of knowledge, but I was wrong. I guess this isn’t the kind of forum where people can ask questions. Why bother wasting everyone’s time asking me why I’m asking a question? I had a question so I asked it. I didn’t ask for negativity, but that’s all you had to offer a future pilot. And I was checkride ready, because I passed! BOOM! Thank you very much for your help. Not!

Congrats on the checkride, kid.

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Help clarify Pilot's difficulty balancing forces acting on aircraft during cruise stage?
How does the ideal disposition of the four forces in an aircraft relate to the nose-up or nose-down pitching moments? What are the difficulties that the pilot can face to maintain such an ideal disposition for each of the forces while cruising? And also, Do these forces values and positions depend on the airspeed and the angle of attack?
THANKS.

It's not real clear what you are asking, but it isn't hard at all.  Once you are all trimmed up in cruise most airplanes are pretty stable in smooth air and fly themselves.

The hard work has already been done by the aircraft designers, making the pilot's job easier.

I suggest you read the section about forces of how it flies. "the 4 forces" is mainly a artificial decomposition of thrust (due to engine), weight and aerodynamic forces (which are sums and then artificially split in its horizontal and vertical component)

… and also the preceding chapters about energy management and angle of attack. Or the whole thing, it's quite good read.

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What is the static thrust of a Piper Arrow III?
I want to find out what's the static thrust of a Piper Arrow III (non-turbo) with Mc Cauley propeller at T/O (max) power as function of the air density.
I can't find it in the manual, I know the engine is rated at 200 HP and had a look at some formulas but they require some other parameters like the propeller efficiency which I can't find.

use 50% efficiency. Then you will be wrong by at most a factor of 2. On a more serious note: The definition of efficiency breaks down at zero speed. A lower value than 50% will be likely. Please see here for some more data.

Weird, according to this http://www.dept.aoe.vt.edu/~lutze/AOE3104/thrustmodels.pdf I should take a 90% "We can estimate the actual static thrust is 90% of that or 1116 lbs."

I'm rather surprised to read that. GA propeller efficiency rarely exceeds 85% and twist is matched to cruise speed, not the static case. Especially fixed-pitch propellers have poor efficiency at zero speed. Here is a NACA report on old measurements of static thrust. See point 5 of the conclusions.

Yes, I read that one as well. Still I must be missing something. If we take a 0.5 for the static thrust we get (with a 6.12 blade) a value of 750 lbs, but then after releasing brakes at 40 knots it would be 814 lbs according to the thrust formula (n * P / v). I can't make sense of it. To me after releasing brakes and speed builds up, the thrust should reduce. What am I missing?

Constant speed or fixed pitch prop?

All Piper Arrows have constant speed props.This model is a 200hp aspirated engine.


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How to calculate aerodrome elevation given QNH and QFE
Would someone be able to explain how to get the answer to the following question?
13) The QFE for the aerodrome is currently 985 hectopascals. The QNH you have been given is 1031 hectopascals. Approximately what would you expect the aerodrome elevation to be?
A 1380 ft
B 690 ft
C 2663 ft
D 875 ft
Correct answer is A.
Thanks

Homework questions are usually off-topic here but I'll wait for the community to decide.

QNH is the pressure setting that will show the aerodrome's true altitude at field level, QFE is the pressure setting where the altimeter will show zero feet at field level. The rule of thumb is that pressure changes 1 hpa per 30 feet, so the formula you would use is (QNH - QFE) * 30feet/hpa = field height. In this case that is (1031hpa - 985hpa) * 30feet/hpa = 46hpa * 30feet/hpa = 1380ft.
If the QFE is greater than QNH you would get a negative number, which would mean the airfield is below sea level. This is pretty rare, but they do exist, so don't mess with the QNH - QFE order.


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Why would a foreign military aircraft be parked at a US Air Force base?
The aircraft in the attached Google Earth screenshot appears to be a Eurofighter Typhoon, parked all alone on the tarmac at Eglin AFB, Florida.  Why would one single foreign fighter like this be found on a US military base?  Granted, the Typhoon is mostly operated by US allies, but it still seems strange.
Also, can anyone identify the livery/operator of the aircraft in question?

As to the operator, it's tough to tell from the low resolution but my guess would be the UK (spanish jets appear to have roundels on both wings, german jets have the cross instead, italians have a red circle roundel, Oman does not appear to put them on either wing, and as seen in this picture, Saudi typhoons have a more subdued roundel and "RSAF" on the right wing). UK jets feature a blue outer circle on their roundel

It is actually fairly common for NATO aircraft to be at Eglin AFB.  When I was stationed there (a long time ago) we had two RAF Tornadoes on site for several months.
Eglin AFB is the home to the munitions development programs. Part of that development includes ensuring the weapons are compatible with the aircraft that will deploy them. Eglin also has a large test range to support that effort.
Since we sell the majority of our munitions throughout NATO (common weapons reduce the  size of stockpiles and increase flexibility), they also need to be tested on NATO aircraft. The Eurofighter is likely there for testing.

It's also very common for NATO (or even non-NATO US allies, like Israeli) military aircraft to be parked at USAF bases during joint exercises such as Red Flag. These are rarely if not never at Eglin, but I don't think OP's question is location-based. So with that said, the premise of the question is very normal, "there's nothing to see here" kind of normal :)


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Could a plane take-off at 20,000 feet above sea level if the temperature is above freezing?
I’ve read that “humans have survived for two years at 5,950 m (19,520 ft, 475 millibars of atmospheric pressure), which is the highest recorded permanently tolerable altitude” and that “Daulat Beg Oldie...has an airstrip at an altitude of 5065 meters (16,614 feet), which is the world's highest airstrip.”  I can’t seem to find any record of the highest horizontal take-off in history.  I really want to know if there is an altitude at which humans can survive but take-off is impossible.  Or if there are any other conditions at which humans could survive naturally but where take-off or flight is impossible for some reason other than temperature, weather or terrain.

6000m is not very high while freezing isn't that cold. I wonder why this could be a problem. If mountaineer used a drone to film the top of Everest he would break the world record?

Are you talking about existing planes or special ones? Design studies have been carried out for planes to fly on Mars, where the air is a heck of a lot thinner and colder.

I should have specified human cargo and not necessarily using existing aircraft.  Are drones and helicopters primarily using vertical take-off instead of horizontal or am I misusing the terms?

The question of if it is possible is moot. I once asked a racecar  builder what was the maximum amount of horsepower you could put in his car. His answer was, “How much money do you have?”. In other words, anything is possible with the right amount of time, resources, and desire.
In the case of your question, it is not about whether an airplane can do it. It is not even about whether a current military, cargo, passenger, jet, prop-driven, or currently produced plane can do it. It is more about where can you do it. You would need the proper facilities (long enough runway). A helicopter has already exceeded your altitude in question due to not having the runway restriction.
Didier Delsalle

Like most hypothetical questions of this nature, the answer is "it depends".  Given enough runway at a light loading just about any aircraft with a service ceiling above 20K ought to theoretically be able to take off.  One of the issues you might run into would be exceeding max tire speed.

Or give the plane a lot of power.  FTM, NASA has built a helicopter that's intended to be flown on Mars, where the density altitude is WAY over 20K ft: https://mars.nasa.gov/technology/helicopter/


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What design characteristics influence helicopter wind gust spread tolerance during takeoff and landing
Not considering pilot reaction time, what factors influence a helicopter's ability to survive high wind gust spreads (from any direction) during takeoff and landing?
I have considered:

Swash plate delay: assuming an example main rotor spinning at 600RPM, the rotor completes 10 rotations per second. If each control input requires 180 degrees of rotation to take effect, then fastest possible reaction time is an adjustment every 1/20 second.
Mass of the aircraft: a heavier aircraft will be moved a shorter distance by crosswind gusts. This reduces the likelihood a skid will unintentionally strike the ground during the earliest stages of takeoff or the final stages of landing (1inch to 3ft off the ground), which could result in the gust building under the vehicle and causing a  roll.
Rotor system: rigid and articulating rotor systems are not susceptible to mast bumping.

Questions:

How do you rank these factors in order of largest influence on surviving high wind gusts during takeoff?
Are there other factors I have not considered?

Swashplate delay is not in rotations. The point of swashplate is to adjust the blade pitch differently according to their position, so the blade pitch must precisely follow the swash plate position. Control delay is because it takes some force, and thus time, to adjust the swash plate (by pilot or by hydraulic actuators if involved) and because the swash plate only sets rate of tilt and the maximum rate of tilt is limited by the available blade pitch range and resulting lift and the rotational inertia.

Thanks @JanHudec - I will further dig into swashplate mechanics

Not an engineer but I would assume disk loading, and rotor tip speed are major players in the game. higher disk loading, and faster tip speeds would likely be much less susceptible to gusts, looking forward to someone who knows what there talking about. Vary Effective tail-rotors are necessary if the gusts are not right on the nose


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Maglev catapult aeroplane takeoff concept
It occurs to me that an aeroplane that is catapulted off a runway via a maglev system would eliminate many current problems we face With aircraft travel. Firstly the runway could be as long as they are now - or shorter if that is better. Secondly a maglev system has unlimited and immediate torque. Any kind of takeoff speed could be provided up to about 500km per hour, perhaps even more. Thirdly, the thrust would be so massive that the engines would not have to be used at maximum thrust for takeoff - so reducing wear and tear, but more importantly aircraft noise and pollution at airports. Fourth, the aircraft wheels would have to be mounted onto maglev ‘Shoes’ which would be heavy of course - but they would return to takeoff position after each takeoff. Fifth, the amount of onboard fuel required would then also be much less, making the aircraft lighter as it climbed. Sixth, the noise levels in the airport would be much lower so eliminating much of the political opposition to new runways - and supporting the case for more night flights. Seventh, as the plane would be mounted on shoes on the maglev and runway distance would be much greater than for steam catapult takeoff the stresses on the aircraft would not be any greater than they are for current takeoffs - And there would no wheel friction/pollution. Finally, eighth, there is plenty of renewable power now available to power the maglev, so we could potentially cut aircraft fuel consumption by almost 30pc, while eliminating aircraft fuel pollution around airports by more than 50pc. I believe this use of maglev tech is the answer to our aircraft-based global warming concerns - and so could save the travel industry. Maglev is proven and reliable technology.

I feel like there is a duplicate question around here somewhere.

Going through such trouble to save 20 seconds of fuel?

"Thirdly, the thrust would be so massive that the engines would not have to be used at maximum thrust for takeoff" Most airliner take-offs are not done at "maximum thrust".

This isn't a question.

This sounds like the musings of someone with just enough knowledge to think outside of the box. But, not enough to recommend feasible solutions. I applaud outside of the box thinking. But, this site is a question and answer forum for factual info. Not for a pro/cons discussion based on nothing but opinion. This question will probably be closed on this site. Might I suggest the world building stack exchange to flush out your argument first.

The Americans can't get the electromagnetic catapults working on their new aircraft carriers See the fuss about the system installed on the Gerald R Ford. This is not a mature technology, and couldn't be employed for passenger transport, even if it could deliver the benefits you claim.

@Dean F, so true, and well said.  I try to be patient, but there are an awful lot of people out there who think they've had some brilliant insight that will revolutionize aviation who don't realize that the technology evolution curve is in that spot where even small incremental improvement comes at great cost.  And who don't seem to understand cost benefit and break even analysis.

Although this might not be an answer to a question, there was no question here.

“It occurs to me that an aeroplane that is catapulted off a runway via a maglev system would eliminate many current problems we face With aircraft travel.”

What problems are we having with current airport facilities?

”Secondly a maglev system has unlimited and immediate torque.”

The maglev system you propose would not provide torque. It would provide force. And, not unlimited force. The more force you need, the bigger the mechanism required, the more money spent. A jetliner is different from a roller coaster car or even a fighter jet.

”Thirdly, the thrust would be so massive that the engines would not have to be used at maximum thrust for takeoff...”

Aircraft would need thrust or force to accelerate to rotation speed. Your system would provide that much. After that, the aircraft would still need to continue producing the same thrust it uses now to accelerate to maximum climb rate or maximum climb angle. A catapult system would stop producing thrust/force immediately upon the airplane leaving the ground. The only thing this would solve would be to shorten the distance necessary to reach rotation speed. For example, fighters leaving aircraft carriers still must use enough thrust to climb out.

”Fifth, the amount of onboard fuel required would then also be much less, making the aircraft lighter as it climbed.”

The difference in fuel consumption rate between takeoff trust with a catapult and takeoff thrust without one would be the same. Turbine engines have to have time to spool up. The aircraft would have to use takeoff/go around power prior to the start of being catapulted regardless. Look at fighter jets being launched from an aircraft carrier. If it were possible for the engines to go from completely off to climb thrust in one second, you would only save less than 60 seconds of fuel.

”Sixth, the noise levels in the airport would be much lower so eliminating much of the political opposition to new runways - and supporting the case for more night flights.”

When looking at aircraft carriers as an example, you would increase noise. This is because you would have to take the engines to takeoff thrust immediately before the launch. More time on the ground would be spent at the maximum noise level.

”Seventh, as the plane would be mounted on shoes on the maglev and runway distance would be much greater than for steam catapult takeoff the stresses on the aircraft would not be any greater than they are for current takeoffs”

????
The stress on the total airframe would not be any greater if you keep the acceleration distance the same. But, the stress would all be put on the front landing gear instead of the engine mounts that are already designed for it.

”Finally, eighth, there is plenty of renewable power now available to power the maglev, so we could potentially cut aircraft fuel consumption by almost 30pc, while eliminating aircraft fuel pollution around airports by more than 50pc.”

???
There is no where near enough renewable energy produced in this world. There is currently not enough renewable energy produced to power non-aviation activities. As mentioned in the above points, this would not decrease the fuel consumption either.
I applaud your out of the box thinking. Unfortunately, your target in this fight to decrease fuel consumption is way too small. You are targeting a phase of flight that takes up less than 60 seconds of flights that can last between one hour and twenty hours. What other benefit can your proposal have besides fuel consumption. Shortening of runways could be a benefit for new and future airports, but not for current ones.

Thankyou for your response and informed critique. In response I would say this is not like short runway catapult. It is a long runway concept. Stresses are much lower and takeoff speed much higher. How much fuel would be saved  if takeoff velocity was 500 km/hr on a commercial airliner? What would be the effect on the engine fans if the plane was being thrust forward by massive ground power? Also you say this would not reduce takeoff noise/pollution, but would it not be more like being fired by a gun - one that accelerates you over a mile or more? Also much wind power is still not stored.

If a bullet weighing 200 tons left the ground at 500 km/hr, what angle of ascent would be possible to achieve the highest point? Obviously the ailerons would have to be able to withstand the stress.

@RichardGillespie - It is less a physics problem and more a practicality problem. In the example of a 200 ton bullet, the gun would be fired at an angle. The greater the angle, the higher the max altitude. Angling the runway would not be practical. The only purpose for your catapult system would be to accelerate the aircraft prior to liftoff. An A380 would leave the ground at 150 knots and continue to accelerate to safely climb away from the ground at 190 knots. It would continue to accelerate to cruise at over 520 knots. These are flying speeds. Below 130 knots is stalling and falling speeds.

@RichardGillespie - Your example of 270 knots (500 kmph) would achieve liftoff. But, the force of drag would decelerate the aircraft far faster than it would for your 200 ton bullet. The engines would have to provide takeoff and climb out thrust while leaving the catapult. Stress on the ailerons would not be a factor. Stress on the passengers would be. This would be from being thrown into the air in a non-linear, less than smooth fashion. Don’t underestimate passenger comfort in the commercial feasibility of this.

@RichardGillespie - On the other end of the spectrum, a Cessna 172SP takes about 3-5 seconds for its engines to reach max horsepower. It takes less than 1000 feet of ground roll (on a good day on a hard surface with low density altitude) to reach the 55 knot speed necessary to rotate the nose to a 10° pitch up attitude. At full throttle, it would continue to accelerate to the 74 knot best rate of climb speed. If thrust is lost after rotation, it will decelerate under its 48 knot stall speed in a matter of 2 or 3 seconds. In a catapult, it would have to maintain full throttle from the ground up


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What would reasonably take to get current to fly general aviation aircraft after many years
What would it take to become current? I was a fairly proficient pilot and accumulated around 5000 hours. In 1984, I changed careers and haven't flown since. I was licensed for commercial and instrument operation of airplanes (single and multi-engine). I had an instrument rating and AGI, IGI, CFIA, CFII and I even  got an ATP certificate (although no type ratings and at 74 years old, I don't think that would be usable). I also have an airframe and power plant mechanic certificate.
For 16 years I lived and breathed aviation. I couldn't make a reasonable living at it so I went back to school and switched to computer science. I am  now retired, still in reasonably good health and I am considering getting back into aviation as a hobby.
I know the rules have changed. Airspace categories are different and airplanes have become incredibly more expensive (holy cow). And the avionics make the stuff I was using look like they should be in a museum.
My question is, "What would it take to make me legal to fly?" Could I ever be a safe pilot again?

Entirely doable.  I have a friend a few years older than you who had less experience and took more time off and he got back into it.  I believe the minimum is a current medical and a flight review.  Of course you will want to get some dual instruction before the review, but how much is entirely up to you.

It's still pretty common to find an aircraft with a 6-pack "steam gauges" with avionics similar to what you would have flown in the 80's. I suggest finding a "rusty pilot" course.

My father is 82 and still flies his Piper Cub all the time. As long as you are still able to pass your flight physical, age is not an obstacle.

I am a private pilot and rent old planes, like 73 Cessna 172, with exactly the old instrument panel.
I am a 74 years old retired civil engineer. and fly only VFR.
Here in our airport, Van Nuy, there are many CFI's who would get you up and going in a few hours and will sign you off when you're ready. they charge $30 per hour; and are much younger than our kids.
you would want to look at an airport near an aviation colleges, those kids need hours as CFI. And are very up to date on the latest technologies and what gear is the most reasonable. Like what navigation on which iPad, headsets, what not.

"...will sign you off when you're ready."  <-this!

Physics is physics.  Planes still fly the same.  It's really a question of book learning.
Another thing, technology is welcome in the cockpit now.  iPads, smartphones, full on glass panels in 172s...it's all fair game.  I personally favor steam gauges but many flight schools offer "modern" technically advanced aircraft (TAA) as rentals now.
I'd guess that many hours of ground will be required along with many hours of home study.  Flying will probably be the easiest thing to "re-learn."
Considering that you are an ATP, the level of knowledge your CFI will require is probably much higher than a VFR-only private pilot returning to the skies.  You will need to know a lot of irrelevant (to you) rules to become thoroughly up to speed again, assuming that you only want to fly recreationally.
Don't let that stop you though.  It's just something to be aware of.

With all due respect, the asker was an ATP decades ago. It's gone now. He hardly aim's to be ATP again. A VFR pilot is a VFR pilot: CFI has no reason or justification to demand more from an ex ATP seeking a VFR licence.

@Jpe61 My school’s standard for a BFR or rental checkout is checkride standards for whatever license you hold, both oral and practical. But there are plenty of CFIs out there who will pencil-whip a BFR if that’s all you want.

Not quite sure what you mean by that, but the asker currently holds no licence. Surely the licence he'll be "reviving" will not be ATP or any other higher lvl stuff. PPL SEP maybe, ME at most. And pencil-whipping anything in aviation tends to have dire effects, sooner or later...

To each his own. If I'm signing my name to an ATP's BFR that pilot will demonstrate ATP knowledge and proficiency. Can you point me to the reg that states his pilot certificates are expired? My understanding of 61.19 seems to conflict with your statement. And the OP most certainly holds pilot certificates. He says so in the question.

But, what about the restriction in FAR 121 that prohibits PIC privileges after age 65? I think it was age 60 when I was flying.

Ok, how about 135?  The point is that OP is an ATP.  I'm not telling anyone what to do, only what I would want to see.  That's all.


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In general, what are the RPM ranges/differences between the various compression stages on a 2 or 3 spool (LP,IP,HP) turbojet/fan engine?
I have a general interest in jet engines and am pretty new to the topic so I wanted to toss this question into the wind as I don't have any previous experience or knowledge to pull from and haven't really found a similarly posted question anywhere else.
In the small amount of reading that I've done when looking at the RPM ranges for the various compressors (LP,IP,HP) generally the LPC spins the slowest and the HPC usually spins the quickest and, if the engine has one, the IPC has an RPM range that falls between the two.
So my question is what sort of speeds or fluctuation in speeds for each shaft could we expect during takeoff/flight/landing? For example does the LPC fluctuate more or less than the HPC/IPC or do they remain fairly constant while in flight?

Related: Which are typical rpm values for aeronautical turbines?

As a rough rule of thumb, the speed of a compressor/fan is such that when at high speed (i.e. takeoff), the tip of the blade is going close to Mach 1 (i.e. the circumferential velocity is close to the local speed of sound).  Thus, large compressors/fan will spin slowly, and small compressors/fans will spin quickly.
If you spin way above Mach 1 (say Mach 2), then you are going to have many problems with shockwaves developing.  But if you are way below Mach 1 (say Mach 0.1), then your compressor is going to be less efficient.  So somewhere around Mach 1 is the sweet spot.  Of course, this will vary with altitude, temperature, forward velocity, etc. Its just a very rough estimate.
For example, a giant engine like a GE9X has a fan speed at takeoff in the 2200 - 2300 rpm range, whereas a medium one like the LEAP is in the 3500-3600 rpm range, and a really tiny one like the Honda HF120 could be 17,000 - 18,000 rpm.
Further, on a given engine the fan is always bigger than the HPC, and thus will spin slower

Mach 1 being the sweet spot makes perfect sense along with smaller diameter stages having higher RPM ranges as the smaller the diameter the smaller the tangential speed is at the blade tip. I assume this is why fighter aircraft in general have smaller size engines so that they can spin faster and provide more thrust if I understand correctly.

fighter engines vs commerical engines are driven by the fact that fighter engines need to go supersonic (here we mean the forward mach number of the plane, not the local circumferential mach number of the blade).  high bypass turbofan do great subsonic but they are not good supersonic.  low bypass turbo is what you want for supersonic speeds. Low bypass fan have lower diameters than high bypass, and thus spin faster.

RPM does not generally fluctuate in flight, or on the ground for that matter.
For 'fan' RPM, 100% N1, is somewhere between 2000 to 4000.(aka LP stage)
For 'core' RPM, 100% N2, is somewhere between 9000 to 14000.(HP stage)
The 3rd spool (IP stage) RPM would be in between the LP and HP nominally around 7000.
These are ballpark figures.

Thank you for the response, I appreciate the ranges even if they are ballpark figures. For airliners I definitely understand why there wouldn't be much if any fluctuation in engine speed for fairly obvious reasons but what if we were to look at a fighter aircraft? Shouldn't we expect there to be more fluctuation?


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Why are private jets more expensive to travel in than commercial aircraft?
I am not sure if this is the right site for a question that asks about the economics of aviation rather than the technical aspects of it, but I couldn't find any other stack exchange site so I am asking this here.
As far as I understand, the reason why commercial aircraft are cheap to travel in is because the operational costs of the aircraft are divided among the many passengers in the flight. A fair share of the maintenance costs come from depreciation of the aircraft and the engine overhaul and inspection. In a business jet, I would assume the cost of fuel and the landing fees are much smaller than that of commercial aircraft and, as a result, do not add much to the cost per seat (since the costs are proportionally smaller - approximately at least). So I suspect that costs such as engine overhaul and aircraft inspection do not scale with aircraft size and are approximately the same for both business jet and large commercial aircraft. As a result, the price that each passenger has to pay is much higher.
I tried looking for a few sources on the internet that confirm this but couldn't find any that show a detailed account of expenses. If this hypothesis is correct, I would appreciate it if you could direct me to a source that compares the operational expenses of any commercial aircraft with a business jet.
So is it safe to say that travelling/ transporting cargo is cheaper in a large aircraft than a smaller one if the fuel efficiency is comparable?

This is a great question, and I would hate to see it closed due to resource questions being off-limits.  I would suggest removing that entire second paragraph.  Anyone answering is free to provide a reference to backup the information they provide.

Aside from the economy of scale effect of a large aircraft vs a small one you can compare apples for apples by looking at the same type used on both roles.
The Canadair CRJ200 was sold as both a regional airliner and a corporate jet, called the Challenger 850, so it makes a good comparative example.
The Challenger 850, being, mechanically, a CRJ200 with a corporate interior, has pretty much the same numbers for fuel burn etc, as a 200.  But it's only carrying 1 to 12 passengers, vs up to 50.  So even "full", its fuel burn per occupied seat mile is always going to be several times higher than the airliner version operating with decent load factors (and they aren't much lighter; the corporate interior is very heavy and with 12 bodies aboard you are close to gross weight with a large fuel load - add in a long range fuel system option by STC for that airplane, and the payload with full fuel was 2 persons... yes, 2).
Then there's utilization.  The CRJ 200 airliner averages about 2500 hours a year.  A Challenger 850, like most corporate jets, averages 300-400 hours per year (corporate shuttle versions fly a lot more, maybe over 1000, but they have regular airline style interiors).  All the fixed costs of operation are amortized over a way smaller number of operating hours (most corporate airplanes would fall apart if you put airliner hours on them; they are not designed for that level of use - this made the Challenger 850, with an airframe meant for 2500 hours/year, absolutely bulletproof in its corporate role).
So you have a cost double whammy.  You're using several times more fuel to get around for each warm body on board (what I might call "warm-body-miles"), compared to an equivalent airliner, and you're doing it infrequently and so are paying through the nose for all the fixed costs like insurance and hangar space, etc.
The reason for the existence for most corporate airplanes is the time valuation of the "warm bodies" travelling on it.  A large company makes the calculation that the value of the CEO, and the CEO's lieutenants' time, justifies the astronomical operating cost of a corporate aircraft; in effect, the jet makes money for the company by making the senior management's job more efficient (in theory - no doubt a lot of it is just "because they can").
If you're the other type of corporate jet owner, the ultrawealthy who own the planes personally, economics doesn't even come into it (I knew a guy who looked after corporate jet purchasers pre-delivery, and he told entertaining stories of bizarre customer meetings, with oligarchs and their children arguing over 500,000 dollar interior options as if they were fighting over who got to lick the cake icing spoon).

Sorry, was out of country and couldn't respond. Great answer though, it clears a lot of things up!


text:
What is a typical range of air temperature used for engine inlet cowl anti icing?
So i was searching the internet for engine inlet cowl anti icing and found that usually compressor bleed air is used to prevent the ice formation in several parts of the aircraft.
I am mainly interested in the inlet cowl anti icing section.
So the compressor bleed air is usually around 450 F ( correct me if i am wrong ).
I assume that this is not the temperature used for anti icing and before it is used for this purpose it is cooled down either because of the pressure difference or it passes through a heat exchanger or something else.
What is a typical range of air temperature used for the cowl anti icing and most importantly what is the lowest limit that is capable to prevent ice formation?
Thank you in advance.

Anti icing systems that use bleed are either "run-wet" or "evaporative".
Run-wet systems heat the surfaces to below the water boiling point, say 140-160F.  Supercooled drops hitting the surface are kept from freezing on the leading edge, but can run back and refreeze on the unprotected areas. In that case the "lowest" temperature is whatever keeps the water liquid until it's off the protected surface.
Evaporative systems heat the surfaces to above the boiling point of water, say 220-230F, so the supercooled drops that strike immediately turn to vapour, like spitting on a steam iron, and no ice can form farther back.  In that case the lowest temperature is the boiling point plus a margin.
You use an evaporative system if you can't tolerate runback ice (otherwise you'll use a cheaper and less demanding run-wet system).  On a wing, this will be related to the airfoil's stall behaviour with runback ice, and a non-slatted wing with a supercritical airfoil (which tends to have poor stall behaviour) will typically have an evaporative system.  If you can live with run-back ice because the wing is tolerant ice forming aft, or if the wing has a slat system that makes it tolerant of ice forming aft with slats out, you may use a run-wet system because of the lower heat demands and costs.
On cowls, same sort of thing; you'll use a run-wet system if the engine is not too sensitive to runback ice forming and distorting the flow on the inside surfaces of the cowl or breaking off and hitting the fan blades, or if it is, an evaporative system, that will keep the cowl leading edges in the mid 200sF.
On the General Electric CF-34, the bleed that supplies the cowl is blended with ambient air to cool it down using an air ejector at the entrance to the cowl piccolo system, where the hot air is dumped into a larger duct from an ejector nozzle, which pulls ambient air in with it and mixes them together. By the time the air is entering the piccolo duct, it exits from little holes cooled off quite a lot, and then it has to fill a space and transfer heat through the skin to the leading edge outer surface, so the it might still be 300-400F (after leaving the bleed port at 500F) or more even when being dumped into the leading edge of the cowl, in order to keep the leading edge skins heated to more than 220F.
If a cowl has a run-wet system, these temps will be much lower obviously.
The upper temperature limit is mostly related to the temperature tolerance of the leading edge and adjacent structure (such as heat treat state and the effects of heating/cooling cycles on the structure etc).

Thank you very much for the explanation!


text:
Where does the air used in an air cooled oil cooler come from in jet engines?
For example, in a turbofan engine the air is taken from the ambient air outside the engine or is it taken from the compressed air from one of the compressor stages ?

Can you give an example for a jet engine with an air cooled oil cooler? The jet engines I'm familiar with use a fuel-oil heat exchanger. I know that some use air cooling for IDG oil. Is that what you mean?

Bleed air, is going to be hot compared to the outside air due to being compressed.  That will make it less effective for cooling purposes.

Kerosene burning high altitude airplanes generally use fuel as the oil cooling medium, because you need to heat the fuel anyway due to the low operating temperatures at altitude. It limits ice formation in the fuel controller, and improves atomization. Sometimes it's just called a fuel heater, not an oil cooler.
If not, then you would use an air/oil heat exchanger, but you still need to heat the fuel, in which case you would have to use bleed air or an electric heater to heat that.
So much simpler to use the hot oil in the engine to warm the fuel. In no case would you use engine bleed, which is hot, to cool engine oil. If it's not an oil/fuel heat exchanger, a stand alone oil cooler will take ambient air ducted to it.

Yes, bleed air would be a bad practice due to its temperature. But if for example the oil temperature is such that the fuel/oil heat exchanger is not enough then and air cooled oil cooler might be used right? Moreover, fuel needs to be heated, as you said to prevent ice particle formation and to protect the filter. In a top of climb case of a civil transport turbofan engine, what would be a reasonable fuel temperature supplied to the fuel oil heat exchanger? Thank you in advance.

A fuel/oil heat exchanger is WAY more efficient than a oil/air heat exchanger.  The supplied fuel will be somewhere above ambient after a long climb, and will cool toward ambient. So if it's -30C, eventually the fuel entering the heater will be close to that.

Adding to that, the fuel after serving its purpose of obtaining the heat loads from the various engine subsystems ends up heated to the burner at an increased temperature. So my next question is: should the system's designer goal be a high burner entry fuel temperature with respect to an upper limit ? The upper temperature limit being the point where self ignition and fuel decomposition takes place. In the literature, I bumped into a temperature of around 120 C for Jet A, would that be reasonable?

Not sure. You'd need to talk to a powerplant specialist with an engine OEM.

@qsct In automotive diesel applications (similar fuel as Jet A) the fuel temperature limit is set by the loss of lubricity and the impact on the high pressure fuel pump. As temperature increases, fuel density drops and power is lost. I suspect a similar factor in jet engines.

Thank you both!


text:
Is there any information regarding the combustion efficiency of LH2 over kerosene?
Taking as reference the airbus A220 it needs around 22000 lt of kerosene to be filled. Is there a rough analogy of how much LH2 would the same aircraft need to operate in similar ranges. Of course there will be huge differences in the overall structure of the engine as well as the tanks to support and handle LH2.
Doing the math with the LH2 properties the 22000 lt of kerosene correspond to 91900 lt of Hydrogen at 21K and a pressure of around 1.2 bar, but that doesn't mean that this is the amount of fuel the modified aircraft would need in terms of the LH2 combustion performance.
If I am missing something obvious here, please do correct me. Thank you in advance.

There are a lot of questions here that address liquid hydrogen and its issues & possibilities. Please see if some of those answers address what you're interested in.

For instance, this answer may address your questions.

In order to combust, a fuel needs to be mixed with air. While kerosene needs first to be sprayed into the air stream to form as small droplets as possible which then will evaporate in the heat of the combustion process, liquid hydrogen only needs a bit of heat to become gaseous and will mix quickly. It therefore will burn more readily and more completely compared to kerosene. Hydrogen-powered engines need shorter combustors, and the heat of the compressed air entering the combustor will be plenty enough to heat it.
To quote from the linked answer:

The downsides, of course, are storage and the small molecular size of hydrogen. It is very hard to contain, and it needs big volumes to store a given amount of chemical energy. Pressurized storage at 200 bars holds only 18 kg/m³ or 45 times less than kerosene. With 142 MJ/kg, hydrogen holds three times as much chemical energy than kerosene, but then the volumetric efficiency of kerosene is still a factor of 13.3 better.

Cryogenic storage swaps pressure for low temperature: Below 33 K and above 13 bar, hydrogen becomes a liquid and storage density increases to 30 kg/m³. Still, cryogenic hydrogen storage needs 4 times the volume of the same amount of energy stored as kerosene, plus the isolation and the energy to cool it down and compress it.

Here it should be added that hydrogen is now regularly stored at 700 bar pressure. However, real gas effects make higher pressure less and less effective: Doubling the pressure from 350 bar to 700 bar will only pack 67% more hydrogen into the same volume.
Storing cryogenic fuel needs isolation if it isn't used up quickly. A rocket will burn its hydrogen within minutes, so if a bit boils off it can be tolerated. An airliner needs to fly for several hours, so keeping the hydrogen from boiling is essential if you want to land at your destination with the engines still running (and enough reserve fuel in the tanks).
So what you are missing is:

Tank weight,
the compression rsp. cooling energy, and
embrittlement prevention. Hydrogen will make the usual metals used in airplane construction brittle, so the long-term strength of tanks and pipes to store and transport hydrogen will be compromised if not designed well for that purpose.

A kilogram of kerosene contains a certain amount of stored chemical potential energy, as does a kilogram of hydrogen. These numbers are listed as heating values of fuels in combustion engineering handbooks, where perfect combustion conditions are assumed.  These numbers allow you to determine how much of one fuel is needed to equal the performance of another fuel, which is the missing information you need here.


text:
Why is there no lift component when calculating rate of climb?
I read this post about how to calculate the rate of climb
How does climb-rate vary with density/pressure altitude?
So in the image, all the forces are drawn. I am wondering, why the rate of climb does not depend on the lift force (or not directly (?) ).
So the forces acting in the vertical direction are: a small drag component, a big lift component, a weight component and a thrust component.
A steady climb would mean, that all forces are in balance - so no acceleration in any direction. But on the other hand in on post it is said "Calculate how much power is needed to overcome drag, and any excess can be used for climbing" - so when there is more force available in one direction - there should be an acceleration in this direction?
I am a bit confused ... is the rate of climb = velocity in the body fixed airframe (so where the airplane is flying to) ? Or is this the vertical velocity in z-direction?
Thank you for your support says
Helmut

Related- may help --   https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 .  Note that a steeper climb angle is associated with a smaller lift force, not a larger lift force. This link should also help answer your question about whether the excess thrust causes an acceleration. This link makes the simplifying assumption of constant air density, IAS, and TAS-- to take things to another level and account for the acceleration required to increase the TAS as the aircraft climbs with constant IAS, see https://aviation.stackexchange.com/a/48855/34686 .

There are several different questions here.   You may be able to delete some of it without invalidating an existing answer, and re-post that content separately.  (Perhaps it's too late now-- )

Power and force are different things. Power is force times velocity.

The calculation is done in the reference frame of the air mass. In the reference frame of the aircraft none of the forces are doing work, and the engine power all goes to accelerating the air, but we don't have formulas for that.

The lift is, by definition, orthogonal to the flight path, so it does not do any work. That's why it does not enter the equations directly (it does indirectly due to reduction of induced drag at high flight path angles).

The “power needed to overcome drag” is needed for overcoming the aerodynamic drag. In climb the “aft” direction is tilted own, so part of gravity is added to drag and the “excess can be used for climbing” means the engine is able to produce the higher thrust needed to balance the sum of them.

Rate of climb is vertical velocity, i.e. velocity in the $z$ direction. But for that derivation to work, you need the velocity in the $x$ (horizontal) direction too.

I agree with most of the answer, but not with the 3rd bullet. In climb the (aerodynamic) drag does not increase. But the weight vector has a component along the path. The steeper the climb, the larger the component of weight in the along path direction. The extra power means the engine is able to produce the higher thrust needed to balance the drag + component of weight. #

Not often discussed, but in a  climb aerodrag may decrease due to lower lift requirement (less efficient prop must take up slack), enough to be included in calcs.

@RobertDiGiovanni -- looks like second bullet point already has a reference to that.  Although it's not only induced drag that is being reduced.  If we hold angle-of-attack and  L/D constant, and we increase power and increase climb angle, we actually end up with a reduced airspeed, and therefore reduced Lift and Drag. That's the only way that the force vector diagram can still balance, even though more of the Weight is opposed by Thrust rather than by Lift in the steeper climb. Also covered in https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476

We certainly would take caution to reduce speed in a climb, but the point has interest in that prop efficiency increases with lower airspeed, as would holding AoA optimal.


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Are the Sudan Airways A300-600 (ST-ATA & ST-ATB) stored in Khartoum?
Does anyone know what happened to the last two A300-600R from Sudan Airways? They should be stored but where or are they already dismantled?

Yes they are still stored in Khartoum, however they require extensive repairs if they want to be put back in service


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What is the correct range of the 787-10?
There are two conflicting values for the 787-10's range; they are:

6,345 NM (11,750 km) -- boeing.com

7,000 NM (12,975 km) -- on the same website few years ago, as well as flightglobal.com

Note: In its early development the value was 6,345 NM (11,750 km), same as #1.
Which is correct and why do the values keep changing?

The range of an aircraft depends on the payload. The value you typically find is the range at typical payload.
You can find the payload/range diagram in the Boeing 787 Airplane Characteristics for Airport Planning document:

As you can see, at maximum zero fuel weight, the 787-10 range is only ~4200 NM. According to Wikipedia, the typical passenger configuration for the 787-10 is 330 seats in a 2-class layout. Assuming 195 lbs per passenger (FAA average adult passenger weight in winter) plus 50 lbs baggage per passenger, gives a typical payload of 80,850 lbs. This would result in a range of ~6000 NM. If not everyone onboard is an adult and not everyone takes the full 50 lbs for baggage, the range can be higher of course. Completely empty, the 787-10 can easily fly over 8500 NM.


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How do jet streams affect the fuel consumption?
I am trying to wrap my head around the effect of the jet stream on energy consumption as well as flight time.
My understanding is that everything that has to with the flying part (vs navigation) depends on airspeed. If a plane is flying westbound it experiences the jet stream as headwind and thus more airspeed and free lift? Are the engines then (almost) only used to move forward (which also increases lift - why isn't the plane climbing even higher then?) and does the jet steam produce more friction than "free" lift is worth? Obviously flying eastbound is faster, does the jet stream push the plane with equal speed forward and the engines are only used to produce the required lift as well as a few hundred km/h extra and thus safe time and energy?
Or does all this boil down to some complicated numerical analysis and there is a net gain in doing this which might not be totally obvious?

" If a plane is flying westbound it experiences the jet stream as headwind and thus more airspeed and free lift? "  --this is a deeply flawed assumption, and thus the question is basically unanswerable.

The "free lift" provided by a headwind is only relevant when there is a headwind relative to the motion of the air mass around the plane, as Bianfable points out in their answer. When the aircraft is at 0KT and wants to be going faster (takeoff), or is at some fast speed but wants to be going 0KT (landing), then the headwind is relevant. At altitude the aircraft only cares about its own movement relative to the air mass, and the fact that the air mass happens to be moving relative to the ground does not affect the aerodynamics.

@randomhead -- re "when there is a headwind relative to the motion of the air mass around the plane" -- does this phrase really make sense?

@quietflyer no it doesn't, not really, but I hope my point comes across. I suppose I could say the headwind is only a factor when it is sufficiently unaccounted for in the plane's current momentum: gusts are gusts, whether at 10AGL or 35,000MSL, but when the plane's airspeed is low and particularly when it is trying to connect with or depart from the ground is when the windspeed makes a difference.

As far as aerodynamics is concerned, there is no jet stream1. The aircraft flies at a fixed airspeed with respect to the surrounding air2. How much thrust it needs, how much lift it experiences and so on, only depends on the airspeed w.r.t. to the surrounding air. How fast that air is moving across the ground (wind speed) is irrelevant for aerodynamics.
The ground speed is however affected by the wind. With a strong headwind, the aerodynamics is the same, but you need longer to reach your destination airport (which is usually fixed on the ground). Likewise, a tailwind will get you to your destination faster. There is no "complicated numerical analysis" required to calculate the effect on ground speed:
$$ \text{Ground Speed} = \text{True Air Speed} \begin{cases}
- & \text{Headwind} \\
+ & \text{Tailwind}
\end{cases} $$
Therefore, fuel flow per time will be identical in jet stream east- and westbound. But the total time for the flight (and therefore the total fuel required) will depend on it. Calculating the extra fuel required is not necessarily so straight-forward, since carrying extra fuel makes the aircraft heavier and therefore consume more fuel per time. A heavier aircraft also has a lower optimum cruise altitude, where the wind could be different.

1 Not entirely true, since aircraft can definitely feel the jet stream when entering, leaving or passing through it. This often causes clear air turbulence.
2 This assumes the aircraft flies at the same fixed airspeed of course, but this is usually the case for jet aircraft flying in the jet stream, which fly at a fixed Mach number. Actual airspeed will depend on flight level and temperature.

Thanks for the equation. The distinction between air speed and ground speed isn't always intuitive if you have only dealt with it for a short time. I now understand that I had some wrong assumptions, this happens if your lectures on this topic are still a few semesters away, and you overthink the question.

Best way to conceptualize it is to just analogize it to a ship navigating ocean currents.  If a ship going 15 knots through the water is riding with a current going 3 knots, it's speed over the ocean bottom is 18 knots (call it a "tail current"); if it's going into current, it's going 12 knots (call it a "head current").  If it's going at 90 degrees to the current, it's a wash,  and its speed over the bottom is the same as its speed in the water, 15 knots.
Normally the ship is in a large mass of water moving in one direction, an ocean current.  But maybe near the edge of that mass of water you might have a small area where the current speeds up to 8 knots, because of some phenomena, maybe a rip tide or something.  If you are in the 8 knot current, your over-the-bottom speed goes to 23 knots if it's going your way, or 7 knots if going against you.  You might call this extra fast local current an "ocean jetstream".
If you were crossing the Atlantic, say from the Caribbean to the UK, in the Gulf Stream and taking advantage of its 1 or 2 knot "tail current" to get you there a bit faster, but you knew that somewhere off to the left or right there was a narrow local current going 4 or 5 times faster than that, you might head over to get that that current to really speed up your trip.  On the way back though, you will avoid that current at all costs.  You will also find that the water at the edge of the fast current can be pretty rough from shearing turbulence.
Airplanes are pretty much the same deal, except the currents are gaseous, the ocean bottom is the ground, the speeds are 20-30 times higher, and instead of rough waves at the edge of the fast current, you have Clear Air Turbulence at the edge of the jet.  The jet streams reside in the little wedge of air on the warm side of frontal boundaries, just under the Tropopause, where the air gets squirted along at 80-140 knots in a little ribbon, like a pipeline, a few miles wide.  If you get in it, you want to be going in the same direction or at least just going across, and will move elsewhere to avoid going against it.

Nice answer. Should be upvoted.

Even the pure "side current" will slow the ship down because it will need to change course in order to sail into the correct direction, slowing it down a bit along the desired track.

Well, I'm talking stead state conditions. In any case, that only applies if there is a lag in the correction so the ship proceeds in a semi teardrop pattern and increases the distance travelled. If the current was 0 at point x and increased to 3 knots directly abeam at point y and I adjusted heading to stay on track in concert with the current change, the effect is negligible. Crossing jet streams, if you are more or less perpendicular, you will see the AP point the nose to the upwind side, than back after you leave the jet.  Thrust untouched, no change in CAS, ground speed while this happens.

There are some excellent answers here already, and I won’t try to compete directly with their explanations.  However, your comment about “free lift” from a headwind was an important clue about your thinking, and I wanted to take a slightly different approach in my answer.  (A little more “touchy-feely” perhaps...)
What is apparent is that you are looking at this from an earth bound frame of reference.  Because a kite, tethered to the earth, DOES extract energy the energy from, and experience wind as “free lift”.  In response it climbs.
I know that intellectually you understand that airplanes are not tethered, but try to find a way to untether your mind from this groundling paradigm.  Aircraft are literally “craft of the air”.  They don’t know or care which direction or how fast the surface of the planet below them might be moving, they only respond to and interact with gravity, plus the aerodynamic effects of movement through the viscous three dimensional medium that contains them - The airmass.
Only the pilot cares about the surface of planet below, and only to the extent that he/she desires to interact with it.  (I.e. navigate to a place on it, touch the wheels to it, avoid hitting protrusions of it, etc.)
No “complicated numerical analysis” is required, just a gentle shift in mindset.  Augmented by thoughtful observation of the world around us.  Do you have a river nearby?  Go to it and quiet your mind... Watch objects float by.  (If nothing is floating by, throw a stick in).  Observe how floating objects are one with the current.
If it is safe to do so, go swimming.  Notice the effort needed just to keep position with a spot on the bank?  Relative to the shore, can you swim faster upstream or downstream?  Do you get any “free” energy swimming upstream as you presumed an airplane might get from flying into a headwind?!
Maybe you don’t even need to go swimming, hopefully you “get it” now, so I won’t belabor any further...  ;)
Good luck with your upcoming lectures.

You are correct that airspeed is key to aerodynamics and flight performance. Whereas groundspeed is key to navigation. Neither really affects fuel consumption rate as far as endurance.
Fuel endurance is measured as fuel quantity over time (ex.gallons or pounds per hour). Fuel endurance is a function of engine power. Fuel Range is measured as fuel quantity over distance (miles per gallon or pounds).  Groundspeed in relation to engine power (or really fuel endurance) will affect fuel consumption rate as far as range. Fuel range and fuel endurance are separate but interrelated quantities of fuel consumption.
An aircraft at a set engine power, altitude, barometric pressure, temperature, humidity, and other atmospheric conditions will travel at the same airspeed regardless of the wind speed or direction. Headwinds, tailwinds, or crosswinds will not affect the aircraft airspeed. Therefore, they will not affect the aircraft fuel endurance. However, they will greatly affect the aircraft groundspeed. Which in turn affects the aircraft fuel range.
Also, for the same or similar airspeeds, tailwinds give aircraft faster groundspeeds, and headwinds give aircraft slower groundspeeds. But, the amount of lift produced is a function of airspeed, and not of groundspeed.
Think of it this way. Aircraft fly in and with the airmass as the airmass itself is in motion. Airspeed is the speed relative to the airmass. If you threw a ball from the back to the front of a bus at 60 mph (airspeed) while the bus was moving forward at 60 mph (groundspeed), the ball would be traveling at 120 mph (groundspeed). On the other hand, if you threw a ball from the front to the back of a bus at 60 mph (airspeed) while the bus was moving forward at 60 mph (groundspeed), the ball would be traveling at 0 mph (groundspeed). In both scenarios, the strength/effort it would take to throw the ball would be the same. In this example, the bus is the jetstream (the airmass). The aircraft is the thrown ball. The person doing the throwing is the engines. And, the fuel flow rate is the strength/effort producing the power.


text:
Does a cyclic in a helicopter essentially control rotor attitude?
So the basics of helicopter control as I understand it, at least, in terms of roll and pitch, are as follows:
A helicopter is caused to move through space by means of tilting the main rotor disc, thereby changing the direction of the rotors lift vector to provide both a vertical lift and a horizontal thrust component in the direction that the rotor disc is tilted.  This change in rotor attitude is affected by means of cyclically feathering the blades as they revolve around the hub, thereby creating an imbalance in lift over the whole of the rotor disc.  This imbalance in lift results in the rotor disk, tilting 90° out of phase with the cyclically feathered blades due to gyroscopic precession.
To prevent continued tilting of the rotor disc, once the disc reaches the same attitude as the swash plate, the blades are no longer cyclically feathered, and the rotor continues to hold this attitude.  The rotor is not capable of exceeding the limits of swash plate during flight operations, though semi rigid rotor designs are capable of striking the mast head during operation if an excess of rotor attitude from the horizontal is achieved.
Am I missing something here?

So when a helicopter rotor spins, each blade also moves relative to the rotor hub?

As I understand it, yes.  Either by cyclic input, coning, flapping, or dragging, each rotor blade is in a constant state of movement during forward flight.

The answer to your questions depend heavily on the specific helicopter you want to know about. On one hand, the red bull aerobatic helicopter has rotorblades that are rigidly held at 90 degrees to the hub, while the huey basically hangs freely under the center of its rotor and can exert no torque on it.  In general, there's no direct mapping from swashplate position to rotor position, but I would be totally unsuprised if there's some class of helicopter popular in germany or western canada or something, with that sort of mapping.

The blade tips always describe a perfect circle at any position of the cyclic, and they change their orbits only while the cyclic control is changing...

@QuadmasterXLII But the so-called 'rigid rotors' are not really rigid... They have no flapping or lead-lag (and sometimes not even feathering) hinges, but the blades do really move in flapping, lead-lag and feathering, because the blade roots work as 'virtual' hinges due to their elasticity... So those rotor heads are not really rigid, but hingeless-articulated...

Although a rigid rotorhead isn’t perfectly rigid due to the flexible elements, it’s still, like, pretty rigid. Here’s a first hand account from an oil rig helicopter pilot who’s flown several varieties of rotorhead:
https://www.quora.com/What-are-rigid-and-articulated-rotor-systems-like-to-fly-compared-to-semi-rigid
Here is a first person narrative

the term feathering seems overused here. swash plate increases or reduces AoA but does not feather blades on purpose at all times. Cyclical feathering of some blades can happen, but it's not the norm. One helicopter climbing vertical and then tilting its pitch attitude slightly to translate forward while still climbing has no blade feathering, yet it tilts about pitch axis. all blades are at positive AoA

Just a couple of clarifications from my side.

thereby changing the direction of the rotors lift vector to provide both a vertical lift and a horizontal thrust component

Technically speaking helicopter rotors generate a thrust (not a lift) which, as you correctly wrote, is tilted according to the manoeuvre which has to be flown.

This imbalance in lift results in the rotor disk tilting 90° out of phase with the cyclically feathered blades

The 90° is the delay between disk tilting and the feathering that causes it. The value in ° of this delay depends mainly on the distance of the flapping hinge from the rotor hub: for a teethering rotor the hinge is exactly by the rotor hub and the delay is 90°; for other rotor constructions the delay is smaller.

due to gyroscopic precession

Albeit similar, the delay is caused by the fact that the feathering of the blade is in resonance with the flapping movement: so you don't have to think about someone pushing the border of a spinning top rather to someone pushing a swing.

To prevent continued tilting of the rotor disc, once the disc reaches the same attitude as the swash plate, the blades are no longer cyclically feathered, and the rotor continues to hold this attitude.

Blades reach the attitude commanded by the swashplate (and in turn by the pilot) after the <90° delay (i.e. some 0.04s for a standard EMS helicopter) and they stay in that attitude as long as the pilot keep pushing on the commands. This is just like any other steering system: you rotate the steering wheel to the right and the car turns right. And just because it enters in a right curve doesn't mean that it stops turning: it just keeps on turning unless you don't rotate the steering wheel back in its central position.

The rotor is not capable of exceeding the limits of swash plate during flight operations

Not clear what you mean here.

semi rigid rotor designs are capable of striking the mast head during operation if an excess of rotor attitude from the horizontal is achieved.

I suppose you mean the mast bumping problem. This is true for teetering rotors and it is due to the limited mechanical space between rotor head, blade roots and mast: at low or negative g manoeuvres the rotor head might tilt so much that it clashes against the mast. Something similar might happen also to non-teetering rotors but only at high negative g and actually with catastrophic outcome: in this case the blades are flapping downward so much that they can hit and cut through the tailboom.

I can't edit the question to correct this, so unfortunately this must be left as a comment: it's "mast bumping".

@AdamHyland: right, thanks


text:
Has anyone ever heard of a Bresthaven Dragon-Fly III?
In James Thurber’s classic short story The Greatest Man In The World, the main character, lowlife hooligan Jack ‘Pal’ Smurch, attains the heights of fame when he flies an airplane around the world without stopping.  The type of airplane which he flies is described by Thurber as a “Bresthaven Dragon-Fly III monoplane” and is a “ridiculous contraption”.  While the story is fiction, certain details, such as Smurch’s takeoff and landing at the now defunct Roosevelt Field on Long Island, NY are real places.  I have never heard of a Bresthaven Dragonfly III aircraft, nor that manufacturer.  I suspect the airplane is fictional, but it might be an older era reference to an actual artifact, similar to calling KJFK Idlewylde Airport, etc. Has anyone ever heard of this aircraft type?

Breasthaven might be a real supplier, but more in the... in-flight entertainment direction.

There is no Bresthaven Dragonfly III, nor is Bresthaven an airplane manufacturer.

That's what I get for reading @therac's comment!


text:
Instrument Approaches which do not have a FAF
Most instrument approaches denote the Final Approach Fix (FAF) as the start of the final approach segment terminating in either the Missed Approach Point (MAP) for non-precision approaches or a Decision Altitude (DA) for a precision approach.  On IAPs the FAF is designated either as a Maltese Cross for non precision approaches or as lightning bolt where the glide slope intercept occurs on a precision approach.  That’s all fine and dandy until I encountered this IAP.

There appears to be no FAF designating the final approach segment.  What does this mean?  Are the FAF, the IAF, and the MAP all coincident?
UPDATE
Someone here suggested that this was because the VOR Rwy 27 has straight in minimums.  However that does not appear to be a general rule.  For instance, take VOR Rwy 01 into KSAV.

This approach has straight in and circling minimums yet clearly has a FAF FREMN defined and located by either a DME or GPS.

The Lakeland approach you show does not have a FAF (Final Approach Fix) because there is no defined point (fix), where you are established inbound and start to descend. In this case, the FAF is given by the FAP (Final Approach Point):

FINAL APPROACH POINT− The point, applicable
only to a nonprecision approach with no depicted
FAF (such as an on airport VOR), where the aircraft
is established inbound on the final approach course
from the procedure turn and where the final approach
descent may be commenced. The FAP serves as the
FAF and identifies the beginning of the final
approach segment.

(FAA Pilot/Controller Glossary)
The chart only says "Remain within 10 NM", so the actual FAP depends on how you fly and cannot be depicted on the chart. Note that the approach does not require DME, so the point AMBAZ cannot be identified by every aircraft (which is why the chart lists minima with and without AMBAZ). It therefore cannot be the FAF. The second approach you show (Savannah) requires DME or TACAN, which means it can have a FAF (FREMN) defined via DME (SAV 7.8).
The MAP (Missed Approach Point) of the Lakeland approach is the LAL VORTAC, which is also the IAF (Initial Approach Fix). The Jeppesen chart for this approach explicitly says "MAP at VOR" and the FAA chart you show has the MAP indication (dotted line with upwards arrow) at the LAL VORTAC.

The Jeppesen chart also shows a point FF27 5.7NM before the runway, where a CDFA (continuous descent final approach) at 2.96° should be started, which is probably the point an FMS would use:

The [FF27] and [RW27] are in square brackets because it has no practical use in the NAS.  It is required, however, for FMS and GPS to correctly fly those approaches.  They are called CNF or computer navigation fix.  The final approach segment starts as soon as you are established on course inbound.


text:
Why are there age limits to becoming a pilot in the armed forces?
This may be more related to the armed forces as opposed to aviation but I have to ask:  Does anyone know why most armed forces put an age limit on being a pilot in the service?  Most set the cutoff at 26-30 years of age depending on the service in question.
One hypothesis I always had on this subject is the large amount of money that it costs to train a military pilot (2-3 million USD from basic flight training through advanced training and assignment to a fleet squadron) combined with a long service commitment (7-10 years once training is completed) where the individual must maintain good health and good eyesight to meet the medical standards required to fly and the mid 30s- mid 40s is when a lot of medical problems begin to crop up and can interfere with this.  As such the service limits the age so they enter this range at the tail end of their military flying careers.  But that's just a guess.
Another is that the competitive nature of military flight slots allow the services to be choose and only select young candidates to weed out the applicant pool.
In the civilian world there is an age limit of 65 for being a commercial airline pilot as the only real restriction to flying commercially but people frequently enter into that career in their 30s and 40s.  It does seem abnormal, if you meet the other medical requirements for the armed forces to limit a career as a pilot considering that other military occupations allow people to enter well into their late 30s or early 40s.  Does anyone have any information on this limit and the reasons why?

See What is the “optimal” age of a fighter pilot?

That's not really an answer to the question.  It just indicates an optimal age people are different and an 'optimal age' varies from person to person.

cutoff at 26-30 you sure? that would make them train the pilots just for a few years of service, not really optimal.

Usually, most services state an age limit to become (not remain) a pilot in the range.  The Army and AF state you must have been commissioned as an officer by the age of 29 and the Navy and USMC set their limit at age 26.  I dunno what USCG requirements are, but I imagine similar.

As always there is a waiver for everything in the military.  Given how desperate the services are to get pilots right now, that could change.

ah, then it's an entry limit, the way you phrased it, it seemed to me you were saying that that is an exit limit (i.e., no matter what, by 30 they will have thrown you out of active duty)

Just an opinion: Younger people learn and memorize more (in general). From a health point of view, younger people recover more completely and quicker, from any traumatism or illness. Add that to better senses (eyesight, hearing) and reflexes, and likely more enthusiasm. Older people have more experience, but that doesn't seem to compensate.

In part, it's probably because it's the military, so things are often done "because that's the way we've always done it", rather than for any rational reason.

"In the civilian world there is an age limit of 65 for being a commercial airline pilot" - 60, not 65.

Largely for the same reason there is an upper bound on how old you can be to start training as an FAA Air Traffic Controller: Training people is expensive and time-consuming (flight training moreso than most military specialties), and we want to make sure we're going to get a certain number of years where the pilot is operating at peak performance out of that investment.
Combat flying - particularly in fighters - is a young man's game: It is physically demanding, and it's much easier to maintain the kind of physical conditioning required when you're younger.
Pilot contracts are also typically longer than other enlistment contracts (I believe the going term these days is 10 years of active service, after you "earn your wings"): A pilot who enters training at the maximum age (currently 28 for the US Air Force) will be 38-39 upon leaving the service, if they re-enlist for another 9 year contract they will be in their late 40s (when the natural process of aging will likely begin to limit the duties they can perform, particularly for more demanding combat roles).
Conversely if they entered at age 35 they would be 45-46 at the end of their first contract, and may already be limited by the natural effects of aging. They would be well into their 50s if they re-enlist, and while we've had active-duty pilots in their 50s they are by far the exception rather than the rule.

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